All categories

3 years ago

The picture shows rays passing through an unknown lens.

Physics

2 answers:

NeX [460]3 years ago

7 0

it id D because it is directed at the middle of the lens so it will be the same

lesya692 [45]3 years ago

7 0

The center point of a lens is called its optical center. According to convention, when the ray of light is parallel to the principal axis, it will pass through the focus and vice versa. Also, when the ray passes through the optical center it will go undeviated.

A ray of light when passes through the optical center of any lens will emerge without any deviation. In case of concave lens, ray D will emerge out.

Similarly, in case of convex lens ray D will emerge out.

So, in either type of lens ray D will emerge out.

You might be interested in

A rifle with a weight of 25 N fires a 4.5-g bullet with a speed of 240 m/s. (a) Find the recoil speed of the rifle. m/s (b) If a

asambeis [7]

Answer:

The recoil speed of the man and rifle is $v_{man}=0.016 ms^{-1}$ .

Explanation:

The expression for the force in terms of mg is as follows;

F=mg

Here, m is the mass and acceleration due to gravity.

Rearrange the expression for mass.

$m=\frac{F}{g}$

Calculate the combined mass of the man and rifle.

$m_{man,rifle}=\frac{650+25}{g}$

Put $g=9.8 ms^{-2}$ .

$m_{man,rifle}=\frac{650+25}{9.8}$

$m_{man,rifle}=68.88 kg$

The expression for the conservation of momentum is as follows as;

$m_{man}u_{man}+m_{bullet}u_{bullet}=m_{man}v_{man}+m_{rifle}v_{man,rifle}$

Here, $m_{man,rifle}$ is the mass of the man and rifle, $m_{rifle}$ is the mass of the rifle, $u_{man},u{bullet}$ are the initial velocities of the man and bullet and $v_{man},v{man,rifle}$ are the final velocities of the man and rifle and rifle.

It is given in the problem that a rifle with a weight of 25 N fires a 4.5-g bullet with a speed of 240 m/s.

Convert mass of rifle from gram to kilogram.

$m_{bullet}=4.5 g$

$m_{bullet}=.0045 kg$

Put $m_{bullet}=.0045 kg$ , $m_{man,rifle}=68.88 kg$ , $u_{man,rifle}=0$ , $v_{bullet}= 240 ms^{-1}$ and $u_{bullet}=0$ .

$m_{man}(0)+m_{bullet}(0)=(68.88)v_{man,rifle}+(.0045)(240)$

$0=(68.88)v_{man,rifle}+(.0045)(240)$

$0=(68.88)v_{man,rifle}+1.08$

$(68.88)v_{man,rifle}=\frac{-1.08}{68.88}$

$v_{man,rifle}=-0.016 ms^{-1}$

Therefore, the recoil speed of the man and rifle is $v_{man}=0.016 ms^{-1}$ .

3 0

3 years ago

It turns out that Mercury and Mars have the same gravity as one another – that is, you would weigh the same on the surface of Me

Tamiku [17]

Answer:

Explanation:

The value of acceleration due to gravity of mars is same as that the value of acceleration due to gravity of Mercury.

The value of acceleration due to gravity of a planet depends on its mass and the radius.

The formula for the acceleration due to gravity is given by

$g=\frac{GM}{R^{2}}$

Where, M be the mass of the planet, R be the radius of the planet.

As according to the question, the value of acceleration due to gravity for Mercury is same as that of Mars.

$\frac{GM_{mercury}}{R_{mercury}^{2}}=\frac{GM_{mars}}{R_{mars}^{2}}$

$\frac{M_{mercury}}{R_{mercury}^{2}}=\frac{M_{mars}}{R_{mars}^{2}}$

As teh radius of Mercury is small, and acceleration due to gravity is same, it is possible because the mass of Mercury is more than the mass of Mars.

4 0

3 years ago

Determine the mass of the object below to the correct degree of precision.

Alinara [238K]

153.73 g.....................................

3 0

3 years ago

a log splitter puts out 1 100W of power for every 1500 W put into it. the efficiency of the machine is what

hram777 [196]

Answer:400 is the answer 1500-1100 is 400

Explanation:

8 0

2 years ago

Read 2 more answers

If x=450 mm, determine the mass of the counterweight s required to balance a 90-kg load, l.

dmitriy555 [2]

<span>The platform scale consists of a combination of third and first class levers so that the load on one lever becomes the effort that moves the next lever.Through this arrangement, a small weight can balance a massive object. If x=450 mm,determine the required mass of the counterweight S required to balance a 90-kg load, l.</span>

3 0

3 years ago